Lecture 02

Psychrometric Processes of Air Conditioning Systems

Version 1 ME, KUETThe specific objectives of this lecture are to:1. Introduction to psychrometric processes and their representation (Section 2.1) 2. Important psychrometric processes namely, sensible and latent cooling and heating,cooling and dehumidification, cooling and humidification, heating and humidification, heatingand dehumidification, chemical dehumidification and mixing of air streams (Section 2.2)3. Representation of the above processes on psychrometric chart and equations for heat andmass transfer rates (Section 2.2) 4. Concept of Sensible Heat Factor, By-pass Factor and Apparatus dew point temperature ofcooling coils (Section 2.2.)5. Principle of air washers and various psychrometric processes that can be performed usingair washers (Section 2.3)6. Purpose of psychrometric calculations (Section 2.4.1)7. Analysis of summer air conditioning system (Section 2. 4.2)8. Selection guidelines for supply air conditions (Section 2.4.3)At the end of the lecture, the student should be able to:1. Represent various psychrometric processes on psychrometric chart2. Perform calculations for various psychrometric processes using the psychrometric chartsand equations3. Define sensible heat factor, by-pass factor, contact factor and apparatus dew pointtemperature4. Describe the principle of an air washer and its practical use5. Estimate the load on the cooling coil and fix the supply conditions for various summerconditioning systems, namely:a) Systems with 100% re-circulation b) Systems with outdoor air for ventilation with zero by-pass factor c) Systems with outdoor air for ventilation with non-zero by-pass factor d) Systemswith reheat for high latent cooling load applications

2.1. Introduction: In the design and analysis of air conditioning plants, the fundamental requirement is to identify thevarious processes being performed on air. Once identified, the processes can be analyzed byapplying the laws of conservation of mass and energy. All these processes can be plotted easily on apsychrometric chart. This is very useful for quick visualization and also for identifying the changestaking place in important properties such as temperature, humidity ratio, enthalpy etc. Theimportant processes that air undergoes in a typical air conditioning plant are discussed below.2.2. Psychrometric processes:Basic Psychrometric processes:1. Sensible cooling, 2. Sensible heating, 3. Latent cooling, 4. Latent heating, 5. Heating andhumidification, 6. Heating and dehumidification, 7. Cooling and humidification, 8. Cooling anddehumidification processes.a) Sensible cooling (process O-A):During this process, the moisture content of air remains constant but its temperaturedecreases as it flows over a cooling coil. For moisture content to remain constant, the surfaceof the cooling coil should be dry and its surface temperature should be greater than the dewpoint temperature of air. If the cooling coil is 100% effective, then the exit temperature of airwill be equal to the coil temperature. However, in practice, the exit air temperature will behigher than the cooling coil temperature. Fig. 2.1 shows the sensible cooling process O-A on apsychrometric chart. The heat transfer rate during this process is given by:Qs = ma(ho – hA ) = ma cp (To – TA) ……………………………………….. (2.1)b) Sensible heating (Process O-B):During this process, the moisture content of air remains constant and its temperatureincreases as it flows over a heating coil. The heat transfer rate during this process is given by: Qs = ma(hB – ho ) = ma cp (TB – To )…………………………………………. (2.2) = ma cpa (TB – To ) + maω cpv (TB – To ) = ma (1.005 + 1.88ω) (TB – To )where cp is the humid specific heat (≈1.0216 kJ/kg dry air) and ma is the mass flow rate of dryair kg/secma = ρQv = ρ.cmm/60 kg d.a./secFor the purpose of air conditioning calculation, standard air is considered at 20oC and 50% RH.Density for standard air is 1.2 kg/m3d.a. and humid specific heat is 1.0216 kJ/kg d.a. K Qs = (cmm)(1.2)(1.0216). ΔT/60 = 0.0204(cmm) ΔT kW……………(2.3)c) Latent Heating or Cooling processes: QL = ma(hB – ho) = ma{(cpTB + hfgoωB) - (cpTo + hfgoωo)} = mahfgo(ωB –ωo) or, QL = (cmm)(1.2)(2500) Δω/60 as latent heat of vaporization hfgo = 2500 kJ/kg d.a. forstandard air = 50(cmm) Δω kW……………………………………………………..(2.4)c) Cooling and dehumidification (Process O-C):When moist air is cooled below its dew-point by bringing it in contact with a cold surface as shownin Fig.2.3, some of the water vapor in the air condenses and leaves the air stream as liquid, as aresult both the temperature and humidity ratio of air decreases as shown. This is the process airundergoes in a typical air conditioning system. Although the actual process path will varydepending upon the type of cold surface, the surface temperature, and flow conditions, forsimplicity the process line is assumed to be a straight line. The heat and mass transfer rates can beexpressed in terms of the initial and final conditions by applying the conservation of mass andconservation of energy equations as given below:By applying mass balance for the water: ma.ωo = ma. ωc + mw ……………………………………………………… (2.3)By applying energy balance: ma.ho = Qt + mw.hw + ma.hc …………………………………………………(2.4)from the above two equations, the load on the cooling coil, Qt is given by: Qt = ma (ho – hc) - ma (ωo – ωc )hw ………………………………….. (2.5)The 2nd term on the RHS of the above equation is normally small compared to the other term,so it can be neglected. Hence, Qt = ma(ho – hc) ……………………………………………………..……… (2.6)It can be observed that the cooling and de-humidification process involves both latent andsensible heat transfer processes, hence, the total, latent, and sensible heat transfer rates (Qt,Ql and Qs) can be written as: Q t = Ql + Q S where Ql = ma (ho – hw) = ma. hfg (ωo – ωc ) Qs = ma (hw – hc) = ma .cp(To – Tc)………………………………. (2.7)By separating the total heat transfer rate from the cooling coil into sensible and latent heattransfer rates, a useful parameter called Sensible Heat Factor (SHF) is defined. SHF is definedas the ratio of sensible to total heat transfer rate, i.e., SHF = Qs /Qt = Qs /(Qs + Ql )…………………………………….. (2.8) From the above equation, one can deduce that a SHF of 1.0 corresponds to no latent heattransfer and a SHF of 0 corresponds to no sensible heat transfer. A SHF of 0.75 to 0.80 is quitecommon in air conditioning systems in a normal dry-climate. A lower value of SHF, say 0.6, implies a high latent heat load such as that occurs in a humidclimate.tan c = Δω/ΔT = 1/2451{(1 – SHF)/SHF}………………………………..(2.9)As (1 – SHF)/SHF=Ql/Qs= 2501x Δω/1.0216x ΔT=2451(Δω/ ΔT)It may be observed from Fig. that point B divides the total enthalpy change (ho – hc) in theratio of SHF and (1 – SHF). The sensible heat transfer taking place along CB is proportional toSHF and the latent heat transfer along BO is proportional to (1 – SHF). The process line OC orCO is called sensible heat factor or process or condition line.We can see that the slope of the cooling and de-humidification or heating and humidificationline is purely a function of the sensible heat factor, SHF. Hence, we can draw the cooling anddehumidification or heating and humidification line on psychrometric chart if the initial stateand the SHF are known. In some standard psychrometric charts, a protractor with differentvalues of SHF is provided. The process line is drawn through the initial state point and inparallel to the given SHF line from the protractor as shown in Fig. 2.4.In Fig.2.3, the temperature Ts is the effective surface temperature of the cooling coil, and isknown as apparatus dew-point (ADP) temperature. In an ideal situation, when all the aircomes in perfect contact with the cooling coil surface, then the exit temperature of air will besame as ADP of the coil. However, in actual case the exit temperature of air will always begreater than the apparatus dew-point temperature due to boundary layer development as airflows over the cooling coil surface and also due to temperature variation along the fins etc.Hence, we can define a by-pass factor (BPF) as: BPF = ( TC – TS)/(TO – TS) similarly it may also be expressed as = (hC –hS)/(hO – hS) = (ωC – ωS)/(ωO – ωS)It can be easily seen that, higher the by-pass factor larger will be the difference between air outlettemperature and the cooling coil temperature. When BPF is 1.0, all the air by-passes the coil andthere will not be any cooling or de-humidification. In practice, the by-pass factor can be increased byincreasing the number of rows in a cooling coil or by decreasing the air velocity or by reducing thefin pitch.Coversely, a contact factor(CF) can be defined which is given by: CF = (1 - BPF)d) Heating and Humidification (Process O-D):During winter it is essential to heat and humidify the room air for comfort. As shown in Fig.2.5, thisis normally done by first sensibly heating the air and then adding water vapour to the air streamthrough steam nozzles as shown in the figure.Mass balance of water vapor for the control volume yields the rate at which steam has to beadded, i.e., mw: mw = ma (ωD – ωO) where, ma is the mass flow rate of dry air From energy balance, Qh= ma (hD – hO) – mwhw where, Qh is the heat supplied through the heating coil and hw is the enthalpy of steam.Since this process also involves simultaneous heat and mass transfer, we can define a sensibleheat factor for the process in a way similar to that of a cooling and dehumidification process.e) Cooling & humidification (Process O-E):As the name implies, during this process, the air temperature drops and its humidityincreases. This process is shown in Fig.2.6, this can be achieved by spraying cool water in theair stream. The temperature of water should be lower than the dry-bulb temperature of airbut higher than its dew-point temperature to avoid condensation.

It can be seen that during this process there is sensible heat transfer from air to water andlatent heat transfer from water to air. Hence, the total heat transfer depends upon the watertemperature. If the temperature of the water sprayed is equal to the wet bulb temperature ofair, then the net transfer rate will be zero as the sensible heat transfer from air to water willbe equal to latent heat transfer from water to air.If the water temperature is greater than WBT, then there will be a net heat transfer fromwater to air. If the water temperature is less than WBT, then the net heat transfer will be fromair to water. Under a special case when the spray water is entirely recirculated and is neitherheated nor cooled, the system is perfectly insulated and the make-up water is supplied atWBT, then at steady-state, the air undergoes an adiabatic saturation process, during which itsWBT remains constant. This is the process of adiabatic saturation discussed in Lecture 1b. Theprocess of cooling and humidification is encountered in a wide variety of devices such asevaporative coolers, cooling towers etc.f) Heating and de-humidification or Chemical dehumidification (Process O-F):This process can be achieved by using a hygroscopic material, which absorbs or adsorbs thewater vapor/moisture from the air. If this process is thermally isolated, then the enthalpy ofair remains constant, as a result the temperature of air increases as its moisture contentdecreases as shown in Fig.2.7 This hygroscopic material can be a solid or a liquid. In general,the absorption of water by the hygroscopic material is an exothermic reaction, as a result heatis released during this process, which is transferred to air and the enthalpy of air increases.g) Mixing of air streams: Mixing of air streams at different states is commonly encountered in many processes,including in air conditioning. Depending upon the state of the individual streams, the mixingprocess can take place with or without condensation of moisture.i) Without condensation: Fig.2.8 shows an adiabatic mixing of two moist air streams during whichno condensation of moisture takes place. As shown in the figure, when two air streams at statepoints 1 and 2 mix, the resulting mixture condition 3 can be obtained from mass and energybalance.From the mass balance of dry air and water vapor: ma,3 ω3 = (ma,1 ω1 + ma,2 ω2) or, ω3 = (ma,1 ω1 + ma,2 ω2)/ma,3From the above equations, it can be observed that the final enthalpy and humidity ratio of mixtureare weighted averages of inlet enthalpies and humidity ratios. A generally valid approximation is thatthe final temperature of the mixture is the weighted average of the inlet temperatures. With thisapproximation, the point on the psychrometric chart representing the mixture lies on a straight lineconnecting the two inlet states. Hence, the ratio of distances on the line, i.e., (1-3)/(2-3) is equal tothe ratio of flow rates ma,2 /ma,1. The resulting error (due to the assumption that the humid specificheats being constant) is usually less than 1 percent.ii) Mixing with condensation:As shown in Fig.2.9, when very cold and dry air mixes with warm air at high relative humidity, theresulting mixture condition may lie in the two-phase region, as a result there will be condensation ofwater vapor and some amount of water will leave the system as liquid water.Due to this, the humidity ratio of the resulting mixture (point 3) will be less than that at point4. Corresponding to this will be an increase in temperature of air due to the release of latentheat of condensation. This process rarely occurs in an air conditioning system, but this is thephenomenon which results in the formation of fog or frost (if the mixture temperature isbelow 0o C). This happens in winter when the cold air near the earth mixes with the humid andwarm air, which develops towards the evening or after rains.2.3. Air Washers:An air washer is a device for conditioning air. As shown in Fig.2.10, in an air washer air comes indirect contact with a spray of water and there will be an exchange of heat and mass (water vapour)between air and water. The outlet condition of air depends upon the temperature of water sprayedin the air washer. Hence, by controlling the water temperature externally, it is possible to control theoutlet conditions of air, which then can be used for air conditioning purposes. In the air washer, themean temperature of water droplets in contact with air decides the direction of heat and masstransfer. As a consequence of the 2nd law, the heat transfer between air and water droplets will be inthe direction of decreasing temperature gradient. Similarly, the mass transfer will be in the directionof decreasing vapor pressure gradient. For example,a) Cooling and dehumidification: Tw < TDPT. Since the exit enthalpy of air is less than its inletvalue, from energy balance it can be shown that there is a transfer of total energy from air towater. Hence to continue the process, water has to be externally cooled. Here both latent andsensible heat transfers are from air to water. This is shown by Process O-A in Fig.2.11.b) Adiabatic saturation: Tw = TWBT. Here the sensible heat transfer from air to water is exactlyequal to latent heat transfer from water to air. Hence, no external cooling or heating of wateris required. That is this is a case of pure water recirculation. This is shown by Process O-B inFig.2.11. This is the process that takes place in a perfectly insulated evaporative cooler.c) Cooling and humidification: TDPT < Tw < TWBT. Here the sensible heat transfer is from air towater and latent heat transfer is from water to air, but the total heat transfer is from air towater, hence, water has to be cooled externally. This is shown by Process O-C in Fig.2.11.d) Cooling and humidification: TWBT < Tw < TDBT. Here the sensible heat transfer is from air towater and latent heat transfer is from water to air, but the total heat transfer is from water toair, hence, water has to be heated externally. This is shown by Process O-D in Fig.2.11. This isthe process that takes place in a cooling tower. The air stream extracts heat from the hot watercoming from the condenser, and the cooled water is sent back to the condenser.e) Heating and humidification: Tw > TDBT. Here both sensible and latent heat transfers are fromwater to air, hence, water has to be heated externally. This is shown by Process O-E in Fig.2.11.Thus, it can be seen that an air washer works as a year-round air conditioning system. Thoughair washer is an extremely useful simple device, it is not commonly used for comfort airconditioning applications due to concerns about health resulting from bacterial or fungalgrowth on the wetted surfaces. However, it can be used in industrial applications.2.4.1. Introduction: Generally from the building specifications, inside and outside design conditions; the latentand sensible cooling or heating loads on a building can be estimated. Normally, depending onthe ventilation requirements of the building, the required outdoor air (fresh air) is specified.The topic of load estimation will be discussed in a later chapter. From known loads on thebuilding and design inside and outside conditions, psychrometric calculations are performed tofind:1. Supply air conditions (air flow rate, DBT, humidity ratio & enthalpy)2. Coil specifications (Latent and sensible loads on coil, coil ADP & BPF)In this chapter fixing of supply air conditions and coil specifications for summer airconditioning systems are discussed. Since the procedure is similar for winter air conditioningsystem, the winter air conditioning systems are not discussed here.2.4.2. Summer air conditioning systems:2.4.2.1. Simple system with 100 % re-circulated air:In this simple system, there is no outside air and the same air is recirculated as shown inFig.2.12 & Fig. 2.13 also shows the process on a psychrometric chart. It can be seen that coldand dry air is supplied to the room and the air that leaves the condition space is assumed to beat the same conditions as that of the conditioned space.The supply air condition should be such that as it flows through the conditioned space it cancounteract the sensible and latent heat transfers taking place from the outside to theconditioned space, so that the space can be maintained at required low temperature andhumidity. Assuming no heat gains in the supply and return ducts and no energy addition dueto fans, and applying energy balance across the room; the Room Sensible Cooling load (Qs,r),Room Latent Cooling Load (Ql,r) and Room Total Cooling load (Qt,r) are given by: Qs,r = ms cpm (Ti - Ts ) ……………………………………………..(2.10) Ql,r = ms hfg (ωi – ωs ) ……………………………………………..(2.11) Qt,r = Qs,r + Ql,r = ms ( hi + hs ) ……………………………………..(2.12) From cooling load calculations, the sensible, latent and total cooling loads on the room areobtained. Hence one can find the Room Sensible Heat Factor (RSHF) from the equation: 𝑄𝑠,𝑟 𝑄𝑠,𝑟 𝑅𝑆𝐻𝐹 = = 𝑄𝑙,𝑟+ 𝑄𝑠,𝑟 𝑄𝑡,𝑟

From the RSHF value one can calculate the slope of the process undergone by the air as itflows through the conditioned space (process s-i) as: 1 1−𝑅𝑆𝐻𝐹 𝑆𝑙𝑜𝑝𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑝𝑟𝑜𝑐𝑒𝑠𝑠 𝑙𝑖𝑛𝑒 𝑠 − 𝑖 , 𝑡𝑎𝑛𝜃 = ……. (2.13) from eqn. 2.9 2451 𝑅𝑆𝐻𝐹Since the condition i is known say, from thermal comfort criteria, knowing the slope, one candraw the process line s-i through i. The intersection of this line with the saturation curve givesthe ADP of the cooling coil as shown in Fig.2.12. It should be noted that for the given roomsensible and latent cooling loads, the supply condition must always lie on this line so that itcan extract the sensible and latent loads on the conditioned space in the requiredproportions.Since the case being considered is one of 100 % re-circulation, the process that the airundergoes as it flows through the cooling coil (i.e. process i-s) will be exactly opposite to theprocess undergone by air as it flows through the room (process s-i). Thus, the temperatureand humidity ratio of air decrease as it flows through the cooling coil and temperature andhumidity ratio increase as air flows through the conditioned space. Assuming no heat transferdue to the ducts and fans, the sensible and latent heat transfer rates at the cooling coil areexactly equal to the sensible and latent heat transfer rates to the conditioned space; i.e., Qs,r = Q s,c & Q l,r = Q l,c ………………………………………… (2.14)The supply condition has to be fixed using Eqns.(2.9) to (2.11). However, since there are 4unknowns (ms, Ts , ωs, and hs ) and 3 equations, (Eqns.(2.9) to (2.11), one parameter has to befixed to find the other three unknown parameters from the three equations.If the by-pass factor (X) of the cooling coil is known, then, from room conditions, coil ADP andby-pass factor, the supply air temperature Ts is obtained using the definition of by-pass factoras: 𝑇𝑠 − 𝑇𝐴𝐷𝑃 𝑋= 𝑇𝑖 −𝑇𝐴𝐷𝑃

or, Ts = TADP + X( Ti – TADP ) …………………………………………….(2.15)

From the mass flow rate of air and condition i, the supply air humidity ratio and enthalpy areobtained using Eqns.(2.10) and (2.11) as: 𝑄𝑙,𝑟 𝜔𝑠 = 𝜔𝑖 − ……………………………………………..(2.17) 𝑚𝑠 ℎ𝑓𝑔

𝑄𝑡,𝑟 ℎ𝑠 = ℎ𝑖 − ………………………………………………….(2.18) 𝑚𝑠

From Eqn.(2.16), it is clear that the required mass flow rate of supply air decreases as the by-pass factor X decreases. In the limiting case when the by-pass factor is zero, the minimumamount of supply air flow rate required is: 𝑄𝑠,𝑟 𝑚𝑠,𝑚𝑖𝑛 = 𝐶 ……………………………………(2.19) 𝑝𝑚 𝑇𝑖 − 𝑇𝐴𝐷𝑃

Thus with 100 % re-circulated air, the room ADP is equal to coil ADP and the load on the coil isequal to the load on the room.2.4.2.2 System with outdoor air for ventilation:In actual air conditioning systems, some amount of outdoor (fresh) air is added to take careof the ventilation requirements. Normally, the required outdoor air for ventilation purposes isknown from the occupancy data and the type of the building (e.g. operation theatres require100% outdoor air). Normally either the quantity of outdoor air required is specified inabsolute values or it is specified as a fraction of the re-circulated air.Fixing of supply condition:Case i) By-pass factor of the cooling coil is zero:Fig.2.13 shows the schematic diagram of the summer air conditioning system with outdoor airand the corresponding process on psychrometric chart, when the by-pass factor X is zero.Since the sensible and latent cooling loads on the conditioned space are assumed to beknown from cooling load calculations, similar to the earlier case, one can draw the processline s-i, from the RSHF and state i. The intersection of this line with the saturation curve givesthe room ADP. As shown on the psychrometric chart, when the by-pass factor is zero, theroom ADP is equal to coil ADP, which in turn is equal to the temperature of the supply air.Hence from the supply temperature one can calculate the required supply air mass flow rate(which is the minimum required as X is zero) using the equation: 𝑄𝑠,𝑟 𝑄𝑠,𝑟𝑚𝑠 = = …………………………………………(2.20) 𝐶𝑝𝑚 𝑇𝑖 −𝑇𝑠 𝐶𝑝𝑚 𝑇𝑖 −𝑇𝐴𝐷𝑃From the supply mass flow rate, one can find the supply air humidity ratio and enthalpy usingEqns.(2.17) and (2.18). ms = mrc + mo ……………………………………………………….(2.21)where mrc is the re-circulated air mass flow rate and mo is the outdoor air mass flow rate.Since either mo or the ratio mo: mrc are specified, one can calculate the amount ofrecirculated air from Eqn.(2.21).Calculation of Coil Loads:From energy balance across the cooling coil; the sensible, latent and total heat transfer rates,Qs,c , Ql,c and Qt,c at the cooling coil are given by: Qs,c = msCpm(Tm – Ts) Ql,c = mshfg(ωm – ωs) Qt,c = Qs,c + Ql,c = ms(hm – hs) …………………………………….(2.22)Where ‘m’ refers to the mixing condition which is a result of the mixing of the recirculated airwith outdoor air. Applying mass and energy balance to the mixing process one can obtain thestate of the mixed air from the equation: 𝑚𝑜 𝜔𝑚 − 𝜔𝑖 ℎ𝑚 − ℎ𝑖 𝑇𝑚 − 𝑇𝑖 = = ≈ ……………………………(2.23) 𝑚𝑠 𝜔𝑜 −𝜔𝑖 ℎ𝑜 − ℎ𝑖 𝑇𝑜 − 𝑇𝑖

Since (mo/ms) > 0, from the above eqn. it is clear that ωm > ωi , hm > hi and tm > Ti. This impliesthat ms(hm - hs) > ms(hi - hs), or the load on the cooling coil is greater than the load on theconditioned space. This is of course due to the fact that during mixing, some amount of hotand humid air is added and the same amount of relative cool and dry air is exhausted (mo =me).From Eqn.(2.10) to (2.12) and (2.22), the difference between the cooling load on the coil andcooling load on the conditioned space can be shown to be equal to: Qs,c – Qs,r = mo cpm (To - Ti ) Ql,c – Ql,r = mo hfg(ωo – ωi) QT,c – QT,r = mo (ho – hi ) ……………………………………….. (2.24) From the above equation it is clear that the difference between cooling coil and conditionedspace increases as the amount of outdoor air (mo) increases and/or the outdoor air becomeshotter and more humid.The line joining the mixed condition ‘m’ with the coil ADP is the process line undergone by theair as it flows through the cooling coil. The slope of this line depends on the Coil Sensible HeatFactor (CSHF) given by: 𝑄𝑠𝑐 𝑄𝑠𝑐 𝐶𝑆𝐻𝐹 = = ……………………………..(2.25) 𝑄𝑠𝑐 + 𝑄𝑙𝑐 𝑄𝑇𝑐

Case ii: Coil by-pass factor, X > 0:

For actual cooling coils, the by-pass factor will be greater than zero, as a result the airtemperature at the exit of the cooling coil will be higher than that of the coil ADP. This isshown in Fig.2.14 along with the process on psychrometric chart. It can be seen from thefigure that when X > 0, the room ADP will be different from the coil ADP. The system shown inFig.2.14 is adequate when the RSHF is high ( > 0.75).Normally in actual systems, either the supply temperature (Ts) or the temperature rise of airas it flows through the conditioned space (Ti-Ts) will be specified. Then the step-wiseprocedure for finding the supply air conditions and the coil loads are as follows:i. Since the supply temperature is specified one can calculate the required supply air flow rateand supply conditions using Eqns. (2.16) to (2.18).ii. Since conditions ‘i’, supply air temperature Ts and RSHF are known, one can draw the line i-s. The intersection of this line with the saturation curve gives the room ADP.iii. Condition of air after mixing (point ‘m’) is obtained from known values of ms and mo usingEqn.(2.23).iv. Now joining points ‘m’ and ‘s’ gives the process line of air as it flows through the coolingcoil. The intersection of this line with the saturation curve gives the coil ADP. It can be seenthat the coil ADP is lower than the room ADP.v. The capacity of the cooling coil is obtained from Eqn.(2.22).vi. From points ‘m’, ‘s’ and coil ADP, the by-pass factor of the cooling coil can be calculated. If the coil ADP and coil by-pass factor are given instead of the supply air temperature, then atrial-and-error method has to be employed to obtain the supply air condition.2.4.2.3. High latent cooling load applications (low RSHF):When the latent load on the building is high due to either too high outside humidity or due tolarge ventilation requirements (e.g. hospitals) or due to high internal latent loads (e.g.presence of kitchen or laundry), then the simple system discussed above leads to very low coilADP. A low coil ADP indicates operation of the refrigeration system at low evaporatortemperatures. Operating the system at low evaporator temperatures decreases the COP of therefrigeration system leading to higher costs. Hence a reheat coil is sometimes used so that thecooling coil can be operated at relatively high ADP, and at the same time the high latent loadcan also be taken care of. Fig.2.15 shows an air conditioning system with reheat coil along withthe psychrometric representation of the process. As shown in the figure, in a system withreheat coil, air is first cooled and dehumidified from point ‘m’ to point ’c’ in the cooling coiland is then reheated sensibly to the required supply temperature ts using the reheat coil. Ifthe supply temperature is specified, then the mass flow rate and state of the supply air andcondition of the air after mixing can be obtained using equations given above. Since theheating process in the reheat coil is sensible, the process line c-s will be horizontal. Thus if thecoil ADP is known, then one can draw the coil condition line and the intersection of this linewith the horizontal line drawn from supply state ‘s’ gives the condition of the air at the exit ofthe cooling coil. From this condition, one can calculate the load on the cooling coil using thesupply mass flow rate and state of air after mixing. The capacity of the reheat coil is thenobtained from energy balance across it.Advantages and disadvantages of reheat coil:a) Refrigeration system can be operated at reasonably high evaporator temperatures leading tohigh COP and low running cost.b) However, mass flow rate of supply air increases due to reduced temperature rise (Ti-Ts ) acrossthe conditioned space.c) Wasteful use of energy as air is first cooled to a lower temperature and then heated. Energy isrequired for both cooling as well as reheating coils. However, this can be partially offset by usingwaste heat such as heat rejected at the condenser for reheating of air.Thus the actual benefit of reheat coil may vary from system to system.4.3. Guidelines for selection of supply state and cooling coil state:

i. As much as possible the supply air quantity should be minimized so that smaller ducts and fans can be usedleading savings in cost of space, material and power. However, the minimum amount should be sufficient toprevent the feeling of stagnation. If the required air flow rate through the cooling coil is insufficient, then it ispossible to mix some amount of re-circulated air with this air so that amount of air supplied to the conditionedspace increases. This merely increases the supply air flow rate, but does not affect sensible and cooling loads onthe conditioned space. Generally, the temperature rise (Ti-Ts) will be in the range of 8 to 15oC.

ii. The cooling coil should have 2 to 6 rows for moderate climate and 6 to 8 rows in hot and humid climate. Theby-pass factor of the coil varies from 0.05 to 0.2. The bypass factor decreases as the number of rows increasesand vice versa. The fin pitch and air velocity should be suitable.

iii. If chilled water is used for cooling and dehumidification, then the coil ADP will be higher than about 4oC.Prob.1. What is the required wattage of an electrical heater that heats 0.1 m3/s of air from15o C and 80% RH to 55o C? The barometric pressure is 101.325 kPa.Soln. Air undergoes sensible heating as it flows through the electrical heaterFrom energy balance, the required heater wattage (W) is given by:W = ma(he−hi) ≈ (Va /νa).cpm(Te−Ti)Where Va is the volumetric flow rate of air in m3/s and νa is the specific volume of dry air. Teand Ti are the exit and inlet temperatures of air and cpm is the average specific heat of moistair (≈1021.6 J/kg.K).Using perfect gas model, the specific volume of dry air is found to be:νa = (Ra.T/pa) = (Ra.T/( pt −pv)) At 15oC and 80% RH, the vapour pressure pv is found to be 1.364kPa using psychrometric chart or equations.Substituting the values of R , T, p and pa Tv in the equation for specific volume, we find thevalue of specific volume to be 0.8274 m3/kg∴ Heater wattage, W ≈ (Va /νa).cpm(Te−Ti)=(0.1/0.8274)x1021.6(55-15) = 4938.8 W (Ans.)Prob.2. 0.2 kg/s of moist air at 45oC (DBT) and 10% RH is mixed with 0.3 kg/s of moist air at25oC and a humidity ratio of 0.018 kgw/kgda in an adiabatic mixing chamber. After mixing, themixed air is heated to a final temperature of 40oC using a heater. Find the temperature andrelative humidity of air after mixing. Find the heat transfer rate in the heater and relativehumidity of air at the exit of heater. Assume the barometric pressure to be 1 atm.Soln. Given: Stream 1: mass flow rate, ma,1 = 0.2 kg/s; T1 = 45oC and RH = 10%.Using psychrometric equations or psychrometric chart, the humidity ratio and enthalpy ofstream 1 are found to be:W1 = 0.006 kgw/kgda & h1 = 61.0 kJ/kgdaStream 2: mass flow rate, ma,2 = 0.3 kg/s; T2 = 25oC and W2 = 0.018 kgw/kgdaUsing psychrometric equations or psychrometric chart, enthalpy of stream 2 is found to be:H2 = 71.0 kJ/kgdaFor the adiabatic mixing process, from mass balance: W3 = (ma,1.w1 + ma,2.w2)/( ma,1 + ma,2)=(0.2.0.006 + 0.3.0.018)/(o.2 + 0.3)= 0.0132kgw /kgdaFrom energy balance (assuming the specific heat of moist air to remain constant): T3 =(ma,1.T1 + ma,2.T2)/( ma,1 + ma,2)=(0.2.45 + 0.3,25)/(0.2+0.3) = 33oCFrom T3 and W3 , the relative humidity of air after mixing is found to be(from chart):RH = 41.8% (Ans.)For the sensible heating process in the heater:Qs = ma(he−hi) ≈ ma.cpm(Te−Ti) = 0.5x1.0216(40-33) = 3.5756 kW (Ans.)The relative humidity at the exit of heater is obtained from the values of DBT (40oC) andhumidity ratio (0.0132 kgw/kgda) using psychrometric chart/equations. This is found to be:RH at 40oC and 0.0132 kgw/kgda = 28.5 % (Ans.)Prob.3. A cooling tower is used for cooling the condenser water of a refrigeration systemhaving a heat rejection rate of 100 kW. In the cooling tower air enters at 35oC (DBT) and 24oC(WBT) and leaves the cooling tower at a DBT of 26oC relative humidity of 95%. What is therequired flow rate of air at the inlet to the cooling tower in m3/s. What is the amount ofmake-up water to be supplied? The temperature of make-up water is at 30oC, at which itsenthalpy (hw) may be taken as 125.4 kJ/kg. Assume the barometric pressure to be 1 atm.Solution: At the inlet to cooling tower: DBT = 35oC and WBT = 24oCFrom psychrometric chart/equations the following values are obtained for the inlet:Humidity ratio, Wi = 0.01426 kgw/kgdaEnthalpy, hi = 71.565 kJ/kgdaSp. volume, νi = 0.89284 m3/kgdaAt the outlet to cooling tower: DBT = 26oC and RH = 95%From psychrometric chart/equations the following values are obtained for the outlet:Humidity ratio, Wo = 0.02025 kgw/kgdaEnthalpy, ho = 77.588 kJ/kgdaFrom mass and energy balance across the cooling tower:Qc = ma{(ho−hi) − (Wo−Wi)hw } = 100 kWSubstituting the values of enthalpy and humidity ratio at the inlet and outlet of cooling towerand enthalpy of make-up water in the above expression, we obtain:ma = 18.97 kg/s, hence, the volumetric flow rate, vi = ma x νi = 16.94 m3/s (Ans.)Amount of make-up water required mw is obtained from mass balance as:m = m (W - Wi) = 18.97(0.02025 − 0.01426) = 0.1136 kg/s = 113.6 gm/s (Ans.)Prob.4. In an air conditioning system air at a flow rate of 2 kg/s enters the cooling coil at 25oCand 50% RH and leaves the cooling coil at 11oC and 90% RH. The apparatus dew point of thecooling coil is 7oC. Find a) The required cooling capacity of the coil, b) Sensible Heat Factor forthe process, and c) By-pass factor of the cooling coil. Assume the barometric pressure to be 1atm. Assume the condensate water to leave the coil at ADP (hw = 29.26 kJ/kg)Soln. At the inlet to the cooling coil; Ti = 25oC and RH = 50%From psychrometric chart; Wi = 0.00988 kgw/kgda and hi = 50.155 kJ/kgdaAt the outlet of the cooling coil; To = 11oC and RH = 90%From psychrometric chart; Wo = 0.00734 kgw/kgda and ho = 29.496 kJ/kgdaa) From mass balance across the cooling coil, the condesate rate, mw is: mw = ma(Wi − Wo) = 2.0(0.00988 − 0.00734) = 0.00508 kg/sFrom energy balance across the cooling tower, the required capacity of the cooling coil, Qc isgiven by: Qc = ma (hi -ho) – mw.hw = 2.0(50.155 − 29.496) − 0.00508 x 29.26 = 41.17 kW (Ans.)b) The sensible heat transfer rate, Qs is given by: Q s = macpm(Ti – To) = 2.0 x 1.0216 x (25 − 11) = 28.605 kWThe latent heat transfer rate, Ql is given by: Ql = mahfg(Wi – Wo) = 2.0 x 2501.0 x (0.00988 − 0.00734) = 12.705 kWThe Sensible Heat Factor (SHF) is given by: SHF = Qs /(Qs + Ql ) = 28.605/(28.605 + 12.705) = 0.692 (Ans.)c) From its definition, the by-pass factor of the coil, BPF is given by:BPF = (To −T ADP)/(Ti − TADP) = (11 − 7)/(25 − 7) = 0.222 (Ans.) The small difference between Qc and (Qs + Ql ) is due to the use of averagevalues for specific heat, cpm and latent heat of vaporization, hfg.